With trig sub it whittlles down nicely.

Let $\displaystyle x=sec({\theta}), \;\ dx=sec({\theta})tan({\theta})d{\theta}$

Make the subs and you get:

$\displaystyle \int_{0}^{\frac{\pi}{3}}tan^{2}({\theta})d{\theta}$

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- Feb 21st 2008, 05:09 PMgalactus
With trig sub it whittlles down nicely.

Let $\displaystyle x=sec({\theta}), \;\ dx=sec({\theta})tan({\theta})d{\theta}$

Make the subs and you get:

$\displaystyle \int_{0}^{\frac{\pi}{3}}tan^{2}({\theta})d{\theta}$ - Feb 21st 2008, 05:12 PMFalconPUNCH!
- Feb 21st 2008, 05:14 PMtopsquark
- Feb 21st 2008, 05:15 PMmr fantastic
- Feb 21st 2008, 05:25 PMmr fantastic
$\displaystyle u^2 = x^2 - 1 \Rightarrow 2u \frac{du}{dx} = 2x \Rightarrow dx = \frac{u}{x} \, du$

$\displaystyle x = 1 \Rightarrow u^2 = 0 \Rightarrow u = 0$. $\displaystyle x = 2 \Rightarrow u^2 = 3 \Rightarrow u = \sqrt{3}$ where the positive root is used.

Then:

$\displaystyle I = \int_{0}^{\sqrt{3}} \frac{u}{x} \frac{u}{x} \, du = \int_{0}^{\sqrt{3}} \frac{u^2}{x^2} \, du = \int_{0}^{\sqrt{3}} \frac{u^2}{u^2 + 1} \, du$

$\displaystyle = \int_{0}^{\sqrt{3}} \frac{(u^2 + 1) - 1}{u^2 + 1} \, du = \int_{0}^{\sqrt{3}} 1 - \frac{1}{u^2 + 1} \, du$,

which integrates easily to give $\displaystyle I = \sqrt{3} - \frac{\pi}{3}$. - Feb 21st 2008, 05:42 PMFalconPUNCH!
Thank you guys for helping me out. I'm going to use this thread as a reference so I won't forget how to do these. (Handshake)

- Feb 21st 2008, 05:50 PMmr fantastic